What is the locus of mid-point of AB? A variable line in a plane passes through a fixed point and meets the coordinate axes at points A and B. What is the locus of mid-point of AB?
What I did is:- 
I took a line passing through a fixed point (a,b) intersecting the Y axis at A and X axis at B. I then took the mid point to be (x,y).
Consequently the points A and B are (0,2y) and (2x,0).
Now I did the following steps:-
$$ \frac{y-b}{x-a} = \frac{2y-0}{0-2x} = - \frac{y}{x} $$ 
On solving it, I got,
$$ \frac{a}{x} + \frac{b}{y} = 2 $$
I have no idea how to proceed further.
Edit:-  The final answer i got i.e $ \frac{a}{x} + \frac{b}{y} = 2 $ is the correct answer and this is an equation of hyperbola itself.
Writing $y$ in terms of $x$ we would get:-
$$ y = \frac{bx}{2x-a} $$ and on plotting a graph (with a,b as constants) we would find that its graph is a hyperbola.
Thank you!
 A: 
More generally, let $p$ and $q$ be two non-parallel lines intersecting at a point $R$.  For a fixed point $S$ on a plane, a variable line $\ell$ passes through $S$, and meets $p$ and $q$ at $A$ and $B$ respectively. For a fixed real number $t$, let $\mathcal{L}$ be the locus of the point $M$ the straight line $AB$ such that $$\overrightarrow{AM}=t\cdot\overrightarrow{AB}.$$  For convenience, $T$ denotes point on the line $PQ$ such that $$\overrightarrow{PT}=t\cdot\overrightarrow{PQ},$$ where $P$ is the intersection of the line parallel to $q$ passing through $S$ with $p$, and $Q$ is the intersection of the line parallel to $p$ passing through $S$ with $q$. 
  
  
*
  
*If $t\notin\{0,1\}$, and if $S$ is not on $p$ or $q$, then $\mathcal{L}$ is equal to hyperbola $\mathcal{H}$ passing through $R$ and $S$, centered at $T$ with asymptotes parallel to $p$ and $q$.
  
*If $t\notin\{0,1\}$, and if $S$ is on $p$ but not on $q$, then $\mathcal{L}$ is equal to the line parallel to $q$ passing through $T$ (although one can argue that $\mathcal{L}$ also includes $p$).
  
*If $t\notin\{0,1\}$, and if $S$ is on $q$ but not on $p$, then $\mathcal{L}$ is equal to the line parallel to $p$ passing through $T$ (although one can argue that $\mathcal{L}$ also includes $q$).
  
*If $t\notin\{0,1\}$, and if $S$ coincides with $R$, then $\mathcal{L}$ has only one element which is $S=R$ (although one can argue that $\mathcal{L}$ is the union of $p$ and $q$).
  
*If $t=0$ and $S\notin p$, then $\mathcal{L}$ is the line $p$.
  
*If $t=0$ and $S\in p$, then $\mathcal{L}$ consists of a single point, namely, $S$ (although one can argue that $\mathcal{L}=p$).
  
*If $t=1$ and $S\notin q$, then $\mathcal{L}$ is the line $q$.
  
*If $t=1$ and $S\in q$, then $\mathcal{L}$ consists of a single point, namely, $S$ (although one can argue that $\mathcal{L}=q$).
  

The seven degenerate cases are obvious.  In this proof, we assume that $t\notin\{0,1\}$, and $S$ does not lie on $p$ or $q$.
Up to an affine transformation, we may assume that $p$ and $q$ are parallel to the horizontal axis and the vertical axis, respectively, and the point $T$ coincides with the origin $O$.  Let $\mathcal{L}$ be the locus of the point $M$.  We want to show that $\mathcal{L}$ is identical to $\mathcal{H}$.  
Without loss of generality, let $S$ have coordinates $(\alpha,\beta)$ wth $\alpha,\beta> 0$.  Note that $\mathcal{H}$ is given by the equation
$$xy=\alpha\beta.$$
Let $U$ and $V$ be the projections of $S$ onto the horizontal axis and the vertical axis, respectively.  Thus, $U=(\alpha,0)$ and $V=(0,\beta)$.  Observe that 
$$R=\left(-\frac{1-t}{t}\alpha,-\frac{t}{1-t}\beta\right).$$ 
Extend $SU$ and $SV$ to meet $p$ and $q$ and $U'=\left(\alpha,-\frac{t}{1-t}\beta\right)$ and $V'=\left(-\frac{1-t}{t}\alpha,\beta\right)$, respectively.
First, we consider the case where $\ell$ has a negative slope.  This means $\ell$ meets the $x$-axis and the $y$-axis at points $C(c,0)$ and $D(0,d)$ with $c,d>0$.  Note that $\triangle CUS\sim \triangle SVD$, so that $$\frac{c-\alpha}{\beta}=\frac{CU}{US}=\frac{SV}{VD}=\frac{\alpha}{d-\beta}.$$
That is, $$(c-\alpha)(d-\beta)=\alpha\beta.$$  Let $A$ and $B$ have the coordinates $\left(r,-\frac{t}{1-t}\beta\right)$ and $\left(-\frac{1-t}{t}\alpha,s\right)$, respectively.   Since $\triangle AU'S\sim \triangle CUS$, we get
$$\frac{r-\alpha}{\beta+\frac{t}{1-t}\beta}=\frac{AU'}{U'S}=\frac{CU}{US}=\frac{c-\alpha}{\beta}.$$
Hence, $r-\alpha=\frac{c-\alpha}{1-t}$.  Similarly, $s-\beta=\frac{d-\beta}{t}$.  Therefore, the point $M$ has coordinates
$$\big((1-t)(r-\alpha),t(s-\beta)\big)=(c-\alpha,d-\beta).$$
Hence, $M$ lies on the hyperbola $\mathcal{H}$.
The case where $\ell$ has a negative slope is done similarly.  Therefore $\mathcal{L}$ is a subset of $\mathcal{H}$.  Conversely, let $M$ be an arbitrary point on $\mathcal{H}$.  Then it is easily seen that the line $\ell$ passing though $SM$ meets $p$ and $q$ at $A$ and $B$ in such a way that $\overrightarrow{AM}=t\cdot\overrightarrow{AB}$ (if $M=S$, then $\ell$ is the tangent to the hyperbola at $S$).  Therefore, $\mathcal{L}=\mathcal{H}$.
A: Let $ A(0,y_A) $ and $ B(x_B,0) $ be an intersect points, $(a,b)$ be a given point and $M(u,v)$ be a point on the locus. 
If $ a=0 $, so $ y=\frac{b}{2} $ is an answer.
If $b=0$, so $x=\frac{a}{2}$ is an answer. 
But for $ab\neq0$ we obtain: $$M\left(\frac{x_B}{2},\frac{y_A}{2}\right),$$
which gives $x_B=2u,$ $y_B=2v,$ $m_{AB}=-\frac{v}{u}$ and an equation of $AB$ it's:
$$y-2v=-\frac{v}{u}x,$$ which gives
$$b-2v=-\frac{v}{u}a,$$ which gives an equation of the locus:
$$bx-2xy=-ay,$$ which is hyperbola
A: Your solution :
Slope of line passing through $(0,2y)$ and $(a,b)$ equals slope of line passing through $(a,b)$ and $(2x,0)$.
$(2y-b)/a=b/(2x-a)$;
$(2y-b)(2x-a)=ab$;
$(y-b/2)(x-a/2)=ab/4$.
A hyperbola (Comment by Aretino).
